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The Carnot cycle is a fundamental concept in thermodynamics that represents an idealized heat engine cycle. It defines the maximum possible efficiency that any heat engine can achieve when operating between two temperature reservoirs. Understanding the Carnot cycle is essential because it directly connects to the Second Law of Thermodynamics and the idea of reversible processes.
Why is the Carnot cycle important? It sets a theoretical benchmark: no real engine can be more efficient than a Carnot engine working between the same two temperatures. This helps engineers and scientists evaluate the performance of practical engines and refrigerators and understand the limitations imposed by nature.
In this section, we will explore the four stages of the Carnot cycle, analyze its efficiency, and see how it applies to real-world systems such as power plants and refrigeration units.
The Carnot cycle consists of four reversible processes involving a working fluid (usually an ideal gas) that undergoes changes in pressure, volume, temperature, and entropy. These processes are:
Let's examine each process in detail.
The working fluid expands slowly while maintaining a constant high temperature \(T_H\) by absorbing heat \(Q_H\) from the hot reservoir. Because the temperature is constant, the internal energy of the fluid remains unchanged. The heat absorbed is converted entirely into work done by the fluid on the surroundings.
Next, the fluid expands without exchanging heat (adiabatically). During this process, the fluid does work on the surroundings, causing its temperature to drop from \(T_H\) to \(T_C\). Since no heat is transferred, the change in internal energy equals the work done.
The fluid is compressed at a constant low temperature \(T_C\), releasing heat \(Q_C\) to the cold reservoir. The surroundings do work on the fluid, and the heat rejected lowers the fluid's volume while keeping its temperature constant.
Finally, the fluid is compressed adiabatically, raising its temperature from \(T_C\) back to \(T_H\) without heat exchange. The work done on the fluid increases its internal energy, preparing it for the next cycle.
Figure: The Carnot cycle shown on Pressure-Volume (P-V) and Temperature-Entropy (T-S) diagrams. The blue curve represents the P-V cycle, while the green rectangle shows the T-S cycle. Heat absorption occurs during isothermal expansion at \(T_H\), and heat rejection during isothermal compression at \(T_C\).
The efficiency of a heat engine is defined as the ratio of net work output to the heat input from the hot reservoir. The Carnot engine, being ideal and reversible, achieves the maximum possible efficiency between two temperatures.
To derive the Carnot efficiency, consider the heat absorbed \(Q_H\) at temperature \(T_H\) and heat rejected \(Q_C\) at temperature \(T_C\). For a reversible cycle, the entropy changes during heat transfer satisfy:
\[\frac{Q_H}{T_H} = \frac{Q_C}{T_C}\]This equality arises because the total entropy change over a reversible cycle is zero.
Rearranging,
\[Q_C = Q_H \frac{T_C}{T_H}\]The net work done by the engine is:
\[W_{net} = Q_H - Q_C = Q_H \left(1 - \frac{T_C}{T_H}\right)\]Therefore, the efficiency \(\eta\) is:
\[\eta = \frac{W_{net}}{Q_H} = 1 - \frac{T_C}{T_H}\]where \(T_H\) and \(T_C\) are absolute temperatures in Kelvin.
This formula shows that efficiency depends only on the temperatures of the hot and cold reservoirs, not on the working fluid or the details of the engine.
| Engine Type | Operating Temperatures (°C) | Typical Efficiency (%) | Carnot Efficiency (%) |
|---|---|---|---|
| Steam Power Plant | Boiler: 550, Condenser: 30 | 35 - 40 | 60 |
| Gas Turbine | Combustor: 1400, Ambient: 25 | 30 - 40 | 80 |
| Automobile Engine | Combustion: 900, Ambient: 30 | 25 - 30 | 70 |
Note: Real engines have efficiencies lower than the Carnot limit due to irreversibilities, friction, and practical design constraints.
Step 1: Convert temperatures to Kelvin.
\( T_H = 550 + 273.15 = 823.15\,K \)
\( T_C = 30 + 273.15 = 303.15\,K \)
Step 2: Use the Carnot efficiency formula:
\[ \eta = 1 - \frac{T_C}{T_H} = 1 - \frac{303.15}{823.15} = 1 - 0.368 = 0.632 \]
Step 3: Convert to percentage:
\( \eta = 63.2\% \)
Answer: The maximum theoretical efficiency of the steam power plant is 63.2%.
Step 1: Calculate the efficiency using the Carnot formula:
\[ \eta = 1 - \frac{T_C}{T_H} = 1 - \frac{300}{600} = 1 - 0.5 = 0.5 \]
Step 2: Calculate net work output:
\[ W_{net} = \eta \times Q_H = 0.5 \times 5000\, \text{kJ} = 2500\, \text{kJ} \]
Answer: The net work output of the Carnot engine is 2500 kJ.
Step 1: Use the COP formula for a Carnot refrigerator:
\[ COP_{ref} = \frac{T_C}{T_H - T_C} = \frac{270}{300 - 270} = \frac{270}{30} = 9 \]
Step 2: Compare with real refrigerator COP:
The real refrigerator has a COP of 3, which is much lower than the ideal COP of 9.
Answer: The Carnot refrigerator has a COP of 9, indicating the real refrigerator operates at about one-third of the ideal efficiency.
Step 1: Calculate entropy change during isothermal expansion:
\[ \Delta S = \frac{Q}{T} = \frac{2000\,J}{500\,K} = 4\, J/K \]
Step 2: For adiabatic expansion, no heat is transferred (\(Q=0\)), so entropy change is:
\[ \Delta S = 0 \]
Answer: Entropy increases by 4 J/K during isothermal expansion and remains constant during adiabatic expansion, consistent with reversible process assumptions.
Step 1: For \(T_H = 400\,K\), \(T_C = 300\,K\):
\[ \eta_1 = 1 - \frac{300}{400} = 1 - 0.75 = 0.25 = 25\% \]
Step 2: For \(T_C = 280\,K\):
\[ \eta_2 = 1 - \frac{280}{400} = 1 - 0.7 = 0.3 = 30\% \]
Answer: Reducing the cold reservoir temperature from 300 K to 280 K increases the Carnot efficiency from 25% to 30%.
When to use: Whenever using formulas involving thermodynamic temperatures such as efficiency or entropy.
When to use: To quickly assess maximum efficiency without detailed fluid properties.
When to use: When analyzing adiabatic expansions or compressions in Carnot or other cycles.
When to use: To better understand process sequences and energy transfers.
When to use: When assessing practical engine or refrigerator performance limits.
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